# Surface Area Problem.

• Nov 30th 2010, 01:11 PM
wair
Surface Area Problem.
In a beehive, each cell is a regular hexagonal prism, open at one end with a trihedral angle at the other end. It is believed that bees form their cells in such a way as to minimize the surface area for a given volume, thus using the least amount of wax in each cell construction. Examination of these cells has shown that the measure of the apex angle theta is amazingly consistent. Based on the geometry of the cell it can be shown that the surface area S is given by

S=6sh-(3/2)s^2*cot(theta)+[(3s^2*(3)^.5)/2]*csc(theta)

where s, the length of the sides of the hexagon, and h, the height, are constants.

a) Calculate ds/d(theta)

b) What angle should the bees prefer?

c) Determine the minimum surface area of the cell in terms of s and h.
• Nov 30th 2010, 01:19 PM
Mondreus
a) Differentiate with regards to $\theta$.
b) Find the angle that gives the smallest surface area (use the result from a).
c) Use the result from b.
• Nov 30th 2010, 01:19 PM
pickslides
Quote:

Originally Posted by wair
a) Calculate ds/d(theta)

do you mean $\frac{ds}{d\theta} or \frac{dS}{d\theta}$ ?
• Nov 30th 2010, 01:44 PM
wair
I'm sorry I mean dS
• Nov 30th 2010, 01:44 PM
wair
Mondreus do I need to do implicit differentiation ?
• Nov 30th 2010, 01:54 PM
Mondreus
No. S is the function and s is simply a constant.
• Nov 30th 2010, 02:10 PM
wair
Oh i see thank you
• Nov 30th 2010, 08:16 PM
wair
OK so for part a I get, dS/d(theta) = (3/2)s^2 * csc^2(theta) - [(3s^2)(3)^.5]/2*csc(theta)cot(theta). can someone confirm if this is right ?
• Nov 30th 2010, 08:57 PM
wair
For part B would I set my derivative to zero and find the min like any max min problem ? And then for part C it says find the minimum surface are in terms of s and h, but my derivative doesn't have an h.
• Nov 30th 2010, 09:19 PM
Mondreus
I will check a) if you post the original problem and your solution in LaTeX code. Your approach to b) is correct. As for c), it is clearly stated that s and h are constants, so you simply exchange $\theta$ for the angle you found in b) (in the S function).
• Nov 30th 2010, 09:33 PM
wair
$S= 6sh-\frac{3}{2}s^2\cot\theta+(\frac{3s^2\sqrt3}{2})\cs c\theta$

$\frac{dS}{d\theta} = \frac{3}{2}s^2\csc^2\theta-(\frac{3s^2\sqrt3}{2})\csc\theta\cot\theta$
• Nov 30th 2010, 10:16 PM
Mondreus
That looks correct. You can also check your answer here: webMathematica Explorations: Step-by-Step Derivatives
• Nov 30th 2010, 10:19 PM
wair
Ok that's good. One more question, for part c I'm assuming I plug in the value of theta that I get into the original surface area function because there is no h term in my derivative. correct ?
• Nov 30th 2010, 10:25 PM
Mondreus
Yes. It's a standard minimization problem.